A 30 cm wrench is used to loosen a bolt with a force applied 0.3 mm from the bolt. It takes 60 N to loosen the bolt when the for

The height measures 69.68 m

Explanation:

Given data

Before striking the ground =  46 % of the total distance

To establish

the height

Solution

We know here acceleration and displacement, which is

d = (0.5)gt²..............1

Here d is the distance, g is the acceleration, and t is time

So, when an object falls it will be

h = 4.9 t²....................2

For the first part of the inquiry

The falling objects account for

54 % of the total distance

0.54 h = 4.9 (t-1)²...................3

Thus,

Now we possess two equations with unknown variables

We can equate both equations

The first equation already solves for h

Substituting h in the second equation allows us to find t

0.54 × 4.9 t² = 4.9 (t-1)²  

t = 0.576 s and  3.771 s

We choose here 3.771 s since 0.576 s is negligible; the distance covered in the last second before it impacts the ground is 46 % of the entire fall.

Thus, selecting t = 3.771 s

Then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h =  69.68 m

Thus, the height is 69.68 m

Answer:

x₂=2×1

Explanation:

According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;

mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.

mgx=(kx)²/2

x=2mg/k----------------compression when the object is at rest

However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy

Thus, if kx²=mv² then

v=x *√(k/m) ----------------where v=0

<pDoubling v results in multiplying x *√(k/m) by 2, leading to x₂ being double x₁

The ideal launch angle of 45° for achieving the greatest horizontal distance is only applicable when the starting height matches the final height. 

<span>In this scenario, you can demonstrate it as follows: </span>

<span>the initial velocity is Vo </span>
<span>the launch angle is α </span>

<span>the initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity becomes </span>
<span>Vh = Vo×cos(α) </span>

<span>the total flight duration is the period required to return to a height of 0 m, thus </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time =? </span>
<span>a = gravitational acceleration = g (= -9.8 m/s²) </span>
<span>therefore </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now let's examine the horizontal distance. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range =? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>therefore </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme points of r (max or min) with respect to α, the first derivative of r with regards to α must be determined and set to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>As Vo and g are constants that are not equal to 0, the only solution for dr/dα to equal 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>